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Klio2033 [76]
2 years ago
9

What adaptation of a cactus protects it from predators? a round cactus with many spines Broad leaves Sharp spines Thick stems Ye

llow flowers
Physics
1 answer:
Alex_Xolod [135]2 years ago
5 0

A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) <u>Collecting dew</u> from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

Learn more about Adaptation of cactus here brainly.com/question/12501143

#SPJ1

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Answer: F

Explanation: The boy is controlling the bowling ball therefore he can use greater force

5 0
3 years ago
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Half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after
makkiz [27]

Answer:

44 years

Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44

3 0
3 years ago
A searchlight is 210 ft from a straight wall. As the beam moves along the​ wall, the angle between the beam and the perpendicula
Ivenika [448]

Answer:

The length of the beam increasing is 9.64 ft/s.

Explanation:

Given that,

Height = 210 ft

Distance =290 ft

According to figure,

We need to calculate the angle

\cos\theta=\dfrac{210}{x}....(I)

Put the value of x in the equation

\cos\theta=\dfrac{210}{290}

\cos\theta=\dfrac{21}{29}=0.72

Now, \sin\theta=\dfrac{20}{29}

On differentiate of equation (I)

-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}

\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}

Put the value in the equation

\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}

\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}

\dfrac{dx}{dt}=9.64\ ft/s

Hence, The length of the beam increasing is 9.64 ft/s.

4 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
4 years ago
A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
4 years ago
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